Let A = {z ∈ C : |z - 2| ≤ 4} and B = {z ∈ C : |z - 2| + |z + 2| = 5}. Then the max {|z1 - z2| : z1

Q: Wait, why can we just assume the maximum distance lies on the real axis? Couldn't it be somewhere else?

That is a brilliant intuition! In general, for arbitrary shapes, the maximum distance could be anywhere. However, because both our disk $A$ and our ellipse $B$ are perfectly symmetric about the real axis, the 'extremes' of these shapes are aligned along this axis. By checking the points furthest apart on this line of symmetry, we capture the global maximum.

Q: How do I know the center of the ellipse is at the origin?

Great question! The definition of an ellipse is $|z - z_1| + |z - z_2| = 2a$. The center is simply the midpoint of the two foci, $z_1$ and $z_2$. Here, $z_1 = 2$ and $z_2 = -2$. The midpoint is $\frac{2 + (-2)}{2} = 0$. So, the center is indeed at the origin $(0,0)$.

Q: Why is the rightmost point of the disk $6$?

The disk is defined by $|z - 2| \le 4$. This means all points $z$ are within a distance of $4$ from the center $2$. On the real axis, the points are $2 - 4$ and $2 + 4$. Thus, the rightmost point is $6$ and the leftmost is $-2$.

Q: Is it possible for the ellipse to be entirely inside the disk?

Excellent check! If the ellipse were entirely inside the disk, the maximum distance would be between the furthest points of the disk itself. However, our ellipse extends to $x = 2.5$ and $x = -2.5$, while the disk extends from $x = -2$ to $x = 6$. Since the ellipse reaches outside the disk on the left, our strategy of picking the extreme points is valid.

Q: Why did we use $2a = 5$ to find the vertices?

In the standard form of an ellipse, $2a$ represents the constant sum of distances from any point on the ellipse to the two foci. This sum is equal to the length of the major axis. Since the major axis lies on the real axis, the vertices are at distance $a$ from the center. Since $2a = 5$, $a = 2.5$, placing the vertices at $2.5$ and $-2.5$.

Question

Let $A = \{z \in \mathbb{C} : |z - 2| \le 4\}$ and $B = \{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\}$. Then the $\max \{|z_1 - z_2| : z_1 \in A 	ext{ and } z_2 \in B\}$ is :

Options:

(A) $\frac{17}{2}$

(B) 8

(C) 9

(D) $\frac{15}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the complex plane! Today, we are not just solving a math problem; we are mapping a landscape.

We have two sets, $A$ and $B$, living in the complex plane $\mathbb{C}$. Our mission is to find the maximum possible distance between any point $z_1 \in A$ and any point $z_2 \in B$.

This is a classic optimization problem that tests your ability to visualize abstract equations as physical, geometric objects.

### Decoding the Disk

Let us first look at set $A$, defined by the inequality $|z - 2| \le 4$. In the language of complex numbers, $|z - z_0| \le r$ is the universal signature of a solid disk.

Here, $z_0 = 2$ is the center, and $r = 4$ is the radius. Imagine standing at the point $2$ on the real axis.

You can walk in any direction, but you cannot go further than $4$ units away. This disk covers the region from $x = -2$ to $x = 6$ on the real axis. It is a stable, circular territory.

### The Elegance of the Ellipse

Now, turn your attention to set $B$, defined by $|z - 2| + |z + 2| = 5$. This is the heartbeat of the problem.

If you recall your conic sections, the locus of points where the sum of distances to two fixed points (the foci) is constant is an ellipse. Here, the foci are at $2$ and $-2$.

The constant sum is $2a = 5$, which tells us that the semi-major axis $a$ is $2.5$. Since the foci are on the real axis, the ellipse is stretched horizontally, centered at the origin $(0,0)$, with its vertices at $2.5$ and $-2.5$.

### The Strategy of Symmetry

We want to maximize $|z_1 - z_2|$. We have a disk $A$ and an ellipse $B$. Both are symmetric about the real axis.

When you have two symmetric shapes, the maximum distance between them almost always lies along the axis of symmetry. If we were to pick a point off the axis, we could always 'rotate' it toward the axis to increase the distance.

Therefore, we focus our search on the real axis. To maximize the distance, we need to pick the point in $A$ that is as far to the right as possible and the point in $B$ that is as far to the left as possible (or vice versa).

### The Final Calculation

The rightmost point of our disk $A$ is found by taking the center $2$ and adding the radius $4$, giving us $z_1 = 6$. The leftmost point of our ellipse $B$ is found by taking the center $0$ and subtracting the semi-major axis $2.5$, giving us $z_2 = -2.5$.

Now, the distance is simply the difference between these two points on the real line:

$$|z_1 - z_2| = |6 - (-2.5)| = |6 + 2.5| = 8.5$$

Converting this to a fraction, we get:

$$8.5 = \frac{17}{2}$$

The **maximum distance** between the two sets is **$\frac{17}{2}$**.

### Reflection

Look at the beauty of this result. By translating algebraic constraints into geometric shapes, we turned a daunting complex number problem into a simple exercise of finding the distance between two points on a line.

Never fear the complex plane; it is just a canvas for your geometric intuition. Keep practicing, keep visualizing, and the math will always reveal its secrets to you.

Animated Solution for Mathematics - Complex Numbers: Let $A = {z \in C : ∣ z - 2∣ \leq 4}$ and $B = {z \in C : ∣ z - 2∣ + ∣ z + 2∣ = 5}$ . Then the $max {∣ z_{1} - z_{2} ∣ : z_{1} \in A and z_{2} \in B}$ is :

Select Answer:

Visualized Solution

The Optimization Problem

Analyzing Set $A$

Analyzing Set $B$

Calculating Ellipse Parameters

Strategy for Maximum Distance

Finding the Extreme Points

Calculating the Final Distance

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Decoding the Disk

The Elegance of the Ellipse

The Strategy of Symmetry

The Final Calculation

Reflection

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